From c062dd23dbce3e2108e1c3d2705fc323f03fdad3 Mon Sep 17 00:00:00 2001
From: Marcello Seri <marcello.seri@gmail.com>
Date: Tue, 19 Dec 2023 22:21:14 +0100
Subject: [PATCH] fixes

Signed-off-by: Marcello Seri <marcello.seri@gmail.com>
---
 4-cotangentbdl.tex     | 23 ++++++++++++-----------
 6-differentiaforms.tex |  4 ++--
 2 files changed, 14 insertions(+), 13 deletions(-)

diff --git a/4-cotangentbdl.tex b/4-cotangentbdl.tex
index 4c98644..3589d10 100644
--- a/4-cotangentbdl.tex
+++ b/4-cotangentbdl.tex
@@ -364,6 +364,7 @@ \section{One-forms and the cotangent bundle}
 \end{exercise}
 
 Exercise~\ref{ex:propdiff} is particularly interesting if we look at it in relation to the pushforward.
+In particular, the first statement in Theorem~\ref{thm:liealgiso} can be rewritten as follows.
 \begin{proposition}
   Let $F:M\to N$ be a diffeomorphism and $X\in\fX(M)$.
   Then, for any $f\in C^\infty(N)$,
@@ -371,17 +372,17 @@ \section{One-forms and the cotangent bundle}
     X(F^* f) = F_*X(f) \circ F.
   \end{equation}
 \end{proposition}
-\begin{proof}
-  Indeed, for any $p\in M$,
-  \begin{align}
-    F_*X(f) \circ F(p) & = ((F_*X) f) (F(p)) = (F_*X)_{F(p)} f                  \\
-                       & = (dF \circ X \circ F^{-1})(F(p)) f = (dF\circ X)(p) f \\
-                       & = dF_p(X_p) f,                                         \\
-    X (F^*f)(p)        & = X(f\circ F)(p) = X_p(f\circ F) = dF_p(X_p) f         \\
-    % &= d (f\circ F) \circ X = df \circ dF \circ X \\
-    % & = df \circ F_* X \circ F = F_* X (f) \circ F.
-  \end{align}
-\end{proof}
+% \begin{proof}
+%   Indeed, for any $p\in M$,
+%   \begin{align}
+%     F_*X(f) \circ F(p) & = ((F_*X) f) (F(p)) = (F_*X)_{F(p)} f                  \\
+%                        & = (dF \circ X \circ F^{-1})(F(p)) f = (dF\circ X)(p) f \\
+%                        & = dF_p(X_p) f,                                         \\
+%     X (F^*f)(p)        & = X(f\circ F)(p) = X_p(f\circ F) = dF_p(X_p) f         \\
+%     % &= d (f\circ F) \circ X = df \circ dF \circ X \\
+%     % & = df \circ F_* X \circ F = F_* X (f) \circ F.
+%   \end{align}
+% \end{proof}
 In this case you often say that the vector fields are $F$-related\footnote{This is a definition that can be properly formalized, but we will not spend any time on it in during the course.} or that they behave naturally: you can either pull back the function $f$ to $M$ or push forward the vector field $X$ to $N$.
 
 \begin{exercise}
diff --git a/6-differentiaforms.tex b/6-differentiaforms.tex
index c8b0efd..f145923 100644
--- a/6-differentiaforms.tex
+++ b/6-differentiaforms.tex
@@ -619,10 +619,10 @@ \section{Exterior derivative}
 \end{exercise}
 
 \begin{exercise}\label{exe:symplectic}
-  Let $V$ a vector space of dimension $k$.
+  Let $V$ be a vector space of dimension $k$.
   A symplectic form on $V$ is an element $\omega\in\Lambda^2(V)$ which is non-degenerate in the sense that $\iota_v(\omega) = 0$ if and only if $v=0$.
   Cf. Definition~\ref{def:metric}.
-  A \emph{symplectic manifold} is a smooth manifold $M$ equipped with a closed differential 2-form $\omega$ such tht $\omega_q$ is a symplectic form on $T_q M$ for every $q\in M$.
+  A \emph{symplectic manifold} is a smooth manifold $M$ equipped with a closed differential 2-form $\omega$ such that $\omega_q$ is a symplectic form on $T_q M$ for every $q\in M$.
   \begin{enumerate}
     \item Prove that if a symplectic form exists, then $k=2n$ for some $n\in\N$, i.e., it must be an even number.
     \item Let $M$ be a smooth manifold. Define a $1$-form $\eta\in\Omega^1(T^*M)$ on the cotangent bundle of $M$ as